?91IENaturally found uranium consists of 99.274% 2381),

Chapter , Problem 91IE

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Naturally found uranium consists of \(99.274 \%^{238} U, 0.720 \%^{235} U\), and \(0.006 \%^{233} U\). As we have seen, \(235 U\) is the isotope that can undergo a nuclear chain reaction. Most of the \(235 U\) used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, \(U F_{6}\)(g). (a) What is the mass of \(U F_{6}\) in a 30.0-L vessel of \(U F_{6}\) at a pressure of 695 torr at 350 K? (b) What is the mass of \(235 U\) in the sample described in part (a)? (c) Now suppose that the \(U F_{6}\) is diffused through a porous barrier and that the change in the ratio of \(235 U\) and \(235 U\) in the diffused gas can be described by Equation 10.23. What is the mass of \(235 U\) in a sample of the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of \({ }^{235} U F_{6}\) in the sample?

Equation transcription:

Text transcription:

99.274 \%^{238} U, 0.720 \%^{235} U

0.006 \%^{233} U

235 U

U F_{6}

{ }^{235} U F_{6}